Question

好的，我正在尝试显示4个随机图像并从数据库中显示它们。随机性似乎有效，但我不明白如何显示4种不同的结果。它只显示4次结果4次。如果您有任何反馈请告诉我，谢谢。

<div class="similar-entrees-section food-tabs">
<h3>Similar entrees</h3>
<?php if ($food['catId']=1):?>
<?php $random=$db->fetchRow("select * from  foods where catId=1 ORDER BY RAND() LIMIT 4"); ?>
<div class="row">

            <div class="col-xs-6 col-sm-3"> 
            <a href="#" class="food-block">
              <div class="food-img"><img src="img/menu-items/<?php echo $random['image'] ; ?>" class="img-responsive" alt=""></div>
              <h5>Cheddar Omelet
                +&nbsp;Chicken Sausage <span class="chilly"></span> </h5>
              </a> 
              </div>

            <div class="col-xs-6 col-sm-3"> 
            <a href="#" class="food-block">
              <div class="food-img"><img src="img/menu-items/<?php echo $random['image'] ; ?>" class="img-responsive" alt=""></div>
              <h5>Gluten Free Breaded 
                Chicken <span class="chilly"></span></h5>
              </a> 
              </div>

            <div class="col-xs-6 col-sm-3"> 
            <a href="#" class="food-block">
              <div class="food-img"><img src="img/menu-items/<?php echo $random['image'] ; ?>" class="img-responsive" alt=""></div>
              <h5>Mesquite Grilled 
                Chicken Breast</h5>
              </a> 
             </div>

            <div class="col-xs-6 col-sm-3"> 
            <a href="#" class="food-block">
              <div class="food-img"><img src="img/menu-items/<?php echo $random['image'] ; ?>" class="img-responsive" alt=""></div>
              <h5>Mesquite Grilled 
                Chicken Breast <span class="chilly"></span></h5>
              </a> 
            </div>


          </div>
<?php endif; ?>

Answer 1

首先，您需要替换

<?php $random=$db->fetchRow("select * from  foods where catId=1 ORDER BY RAND() LIMIT 4"); ?>

带

<?php 
    $res = $db->query("select * from  foods where catId=1 ORDER BY RAND() LIMIT 4");
    $images = $db->getAll($res);
?>

之后，在每张图片之前，而不是

<?php echo $random['image'] ; ?>

你应该使用

<?php echo $images[0]['image'] ; ?>

表示第一张图片

<?php echo $images[1]['image'] ; ?>

第二个......等等。

我还建议你使用html代码为每个图像设置一个循环，最终生成所有图像的html。看看如何循环结果的以下链接。

https://pear.php.net/manual/en/package.database.mdb.intro-fetch.php

Answer 2

fetchRow应该排成一排，而不是一切？

然后你使用<?php echo $random['image'] ; ?>，它会一次又一次地显示同一行。

您可能希望使用类似getAll()（如果您使用的是PEAR）来获取所有结果然后循环它们。

也许就像这样

<div class="similar-entrees-section food-tabs">
    <h3>Similar entrees</h3>
    <?php if ($food['catId'] = 1): ?>
    <?php $random = $db->getAll("select * from foods where catId = 1 ORDER BY RAND() LIMIT 4"); ?>
    <div class="row">
        <?php foreach ($random as $r) { ?>
            <div class="col-xs-6 col-sm-3">
                <a href="#" class="food-block">
                    <div class="food-img">
                        <img src="img/menu-items/<?php echo $r['image']; ?>" class="img-responsive" alt=""></div>
                    <h5>Cheddar Omelet +&nbsp;Chicken Sausage <span class="chilly"></span></h5>
                </a>
            </div>
        <?php } ?>
    </div>
<?php endif; ?>

如何用PHP显示来自MySQl的随机数据

2 个答案: